1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Varvara68 [4.7K]
2 years ago
5

A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

rpendicular to the rod. Two small rings, each with mass 0.200 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.20×10−2 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0 rev/min . Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.
(a) What is the angular speed of the system at the instant when the rings reach the ends of the rod?

(b) What is the angular speed of the rod after the rings leave it?
Physics
1 answer:
Alex73 [517]2 years ago
7 0

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

I_R=\frac{1}{12}ML^2

where

M=3.30\cdot 10^{-2} kg is its mass

L = 0.450 m is its length

Substituting,

I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2

The moment of inertia of the two rings at the beginning is

I_r = 2mr^2

where

m = 0.200 kg is the mass of each ring

r=5.20\cdot 10^{-2} m is their distance from the center of the rod

Substituting,

I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2

So the total moment of inertia at the beginning is

I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2

The initial angular velocity of the system is

\omega_1 = 35.0 rev/min

The angular momentum must be conserved, so we can write:

L=I_1 \omega_1 = I_2 \omega_2 (1)

where I_2 is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

r=\frac{0.450 m}{2}=0.225 m

so the moment of inertia of the rings is

I_r=2(0.200)(0.225)^2=0.0203 kg m^2

and the total moment of inertia is

I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2

Substituting into (1), we find the final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

I_2 = I_R = 5.57\cdot 10^{-4}kg m^2

So using again equation of conservation of the angular momentum:

L=I_1 \omega_1 = I_2 \omega_2

We find the new final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min

You might be interested in
Which color of the visible light waves has the highest energy level?
sveticcg [70]
Red..................
3 0
2 years ago
Read 2 more answers
A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}

3 0
3 years ago
Fernanda is working on a physics project. she is experimenting by rolling a marble down a ramp and then seeing how far it rolls
Ratling [72]

The distance covered on the floor after leaving the ramp is the dependent variable.

  • As a result of the marble's size, the substance it is constructed of, and the angle at which it is placed onto the ground, the distance it rolls varies.
  • Therefore, the angle at which the marble is released onto the ground, the type of material used to make the stone, or its size can all be considered independent variables.
<h3>What is Independent variable?</h3>
  • There are independent and dependent variables in every experiment.
  • A variable is considered independent if its change is not influenced by the change in another variable or factor.
<h3>What is Dependent variable?</h3>

In any experiment, the dependent variable must be measured or determined, and it must change as the independent variable does.

Learn more about independent and dependent variable here:

brainly.com/question/1479694

#SPJ4

5 0
2 years ago
A baby born more than 3 weeks before the due date is considered
natulia [17]
Fetal because you are still in the womb and not fully developed
6 0
2 years ago
Read 2 more answers
For a steady two-dimensional flow, identify the boundary layer approximations.
Georgia [21]

Answer:

  • The velocity component in the flow direction is much larger than that in the normal direction ( A )
  • The temperature and velocity gradients normal to the flow are much greater than those along the flow direction ( b )

Explanation:

For a steady two-dimensional flow the boundary layer approximations are The velocity component in the flow direction is much larger than that in the normal direction and The temperature and velocity gradients normal to the flow are much greater than those along the flow direction

assuming Vx ⇒ V∞ ⇒ U and Vy ⇒ u from continuity equation we know that

Vy << Vx

4 0
3 years ago
Other questions:
  • Pls help me solve this physics question​
    12·2 answers
  • How can a budget help you set spending goals?
    7·1 answer
  • Frank does 2400J of work in climbing a set of stairs. If he does the work in 6
    8·1 answer
  • . Imagine that you are standing at the center of a giant bowl of gelatin. What type of wave will you make across the top of the
    5·1 answer
  • Using a 679 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t
    9·1 answer
  • A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point. . . Whats the bul
    14·1 answer
  • A girl and her bicycle have a total mads of 42 kg. At the top of the hill her speed is 4 m/s. The hill is 14.3m high and 112m lo
    5·1 answer
  • A particle of ink in a ink-jet printer carries a charge of -8x 10^-13 C and is deflected onto paper force of 3.2x10^-4. Find the
    13·1 answer
  • When a bow releases an arrow, the arrow
    13·2 answers
  • Magnet A has twice the magnetic field strength of magnet B and pulls on magnet B with a force of 100 N. The amount of force that
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!